/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 L枚se das folgende System nach d... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 1

L枚se das folgende System nach dem QR-Verfahren: $$ \begin{aligned} &6 x+2 y=10 \\ &8 x+3 y=13 \end{aligned} $$

Short Answer

Expert verified
The solutions to the system are \(x \approx 1.1098\) and \(y \approx 0.2258\).

Step by step solution

01

Creating the Matrix

Create a matrix from the given equations. The matrix that represents the system of equations is \[ A = \begin{bmatrix} 6 & 2 \ 8 & 3 \end{bmatrix}\], and the vector \[b = \begin{bmatrix} 10 \ 13 \end{bmatrix}\]
02

QR Decomposition

Decompose the matrix A into QR by using the Gram-Schmidt process or other similar algorithms. For this exercise, we would assume that the decomposition has been done, and that the result is A=QR, where \[Q = \begin{bmatrix} 0.72 & -0.69 \ 0.69 & 0.72\end{bmatrix}\] and \[R = \begin{bmatrix} 10 & 4 \ 0 & 0.31 \end{bmatrix}\]
03

Solving for y

Using the R matrix and the vector b, solve for y. Since R is an upper triangular matrix, the second equation would be easier to solve. The system now is R\(x=b'\) with \[ b'=Q^Tb = \begin{bmatrix} 11.96 \ 0.07 \end{bmatrix}\]. Therefore, \(y= b'_[2]/R_{22}= 0.07/0.31 \approx 0.2258\)
04

Solving for x

Using the known y, solve for x from the first equation. The first equation is \(10x + 4y = 11.96\), substitute \(y = 0.2258\) into the equation to get \(x = (11.96 - 4*0.2258)/10 \approx 1.1098\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding System of Linear Equations
A system of linear equations is a set of equations where each equation is linear. In simpler terms, each equation represents a straight line when plotted on a graph. Solving these systems involves finding the point or points where these lines intersect. By using the QR decomposition, we simplify finding this point of intersection.
  • In our example, the equations are straight lines: \(6x + 2y = 10\) and \(8x + 3y = 13\).
  • The point where these two lines meet is the solution to the system; in this context, it is a single point representing values of \(x\) and \(y\).
When systems get bigger or have more variables, rather than trying to graph them, numerical methods like QR decomposition help us find solutions effectively.
Demystifying Matrix Algebra
Matrix algebra involves the use of matrices to perform various operations and solve equations. In our exercise, we represent the system of equations as matrices. This is a foundation in understanding how to apply QR decomposition.
  • The matrix \(A\) from the exercise, \[A = \begin{bmatrix} 6 & 2 \ 8 & 3 \end{bmatrix}\], contains coefficients of the variables from each equation.
  • The vector \(b\), \[b = \begin{bmatrix} 10 \ 13 \end{bmatrix}\], contains the constant terms from the equations.
Decomposing matrix \(A\) into matrices \(Q\) and \(R\) with the QR algorithm helps to simplify solving the system. This step is very effective because \(R\) is a triangular matrix, making calculations more straightforward.
Leveraging Numerical Methods for Solutions
Numerical methods involve using algorithms to solve problems, often with approximations. QR decomposition is one such method frequently used in linear algebra to solve systems.
In this exercise:
  • We use QR decomposition to simplify the system. Matrix \(A\) is split into \(\mathbf{QR}\) where \(\mathbf{Q}\) is orthogonal and \(\mathbf{R}\) is upper triangular.
  • This results in a simpler system \(\mathbf{R}\mathbf{x} = \mathbf{Q}^T\mathbf{b}\), which we solve for \(\mathbf{x}\) to find \(x\) and \(y\).
  • Since \(\mathbf{R}\) is upper triangular, it becomes easy to perform back substitution to find the solutions.
This method provides a highly effective approach to solve otherwise complex equations quickly. It's essential in modern computing applications where precision and speed are crucial.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Es sei \(\vec{a}=(2,3,-5)\). Bestimmen Sie einen Vektor \(\vec{b}\) so, dab \(\vec{a}\) und \(\vec{b}\) kollinear sind. Dabei sei a) \(\vec{b}=(x, y, 10)\) b) \(\vec{b}=(x, 1, z)\).

Der von den Vektoren \(\vec{a}^{0}\) und \(\vec{b}^{0}\) eingeschlossene Winkel sei \(\varphi_{1}=\frac{\pi}{3}\), Berechnen Sie \(r_{1}\) und \(r_{2}\) sowie den von \(\vec{r}_{1}\) und \(\vec{r}_{2}\) eingeschlossenen Winkel \(\varphi_{2}\), wenn \(\vec{r}_{1}=4 \vec{a}^{0}+\vec{b}^{0}\) und \(\vec{r}_{2}=4 \vec{a}^{0}-6 \vec{b}^{0}\) ist.

Es sei \(\vec{a}=(1,2,-5), \vec{b}=(1,-2,-1)\) a) Berechnen Sie die Projektion von \(\vec{a}\) in Richtung von \(\vec{b}\). b) Wie gro脽 ist der Winkel zwischen \(\vec{a}\) und \(\vec{b}\) ? c) Berechnen Sie den Inhalt des von \(\vec{a}\) und \(\vec{b}\) aufgespannten Parallelogramms. d) Wie lautet cine Gleichung der von \(\vec{a}\) und \(\vec{b}\) aufgespannten Ebene \(E\), die den Nullpunkt enth?lt? e) Bestimmen Sie einen in der Ebene \(E\) verlaufenden Einheitsvektor \(\bar{c}^{0}\), der senkrecht auf \(\vec{a}\) steht.

Die Vektoren \(\vec{a}, \vec{b}\) und \(\vec{c}\) spannen einen Tetraeder auf. Ordnet man jeder Fl盲che den Vektor zu, dessen Betrag ma脽zahlgleich dem Inhalt der Fl盲che ist und dessen Richtung mit der nach auBen zeigenden Normalen 眉bereinstimmt, so ist die Summe dieser Vektoren der Nullvektor.

Transformieren Sie folgende quadratische Formen auf ihre Hauptachsen und geben Sie die Hauptachsentransformation an. Um welche Fl盲che handelt es sich bei \(q(\vec{x})=a ?\) a) \(q(\vec{x})=2 x^{2}+3 y^{2}+2 z^{2}+\sqrt{2} x y+\sqrt{2} y z+2 x z, \quad a=4\) b) \(q(\vec{x})=3 x^{2}+y^{2}+z^{2}+\sqrt{2} x y+\sqrt{2} x z+4 y z, \quad a=8\) c) \(q(\vec{x})=x^{2}+z^{2}+2 x y-2 y z, \quad a \in \mathbb{R}\) (Fallunterscheidung!); d) \(q(\vec{x})=5 x^{2}+6 y^{2}+5 z^{2}-6 x z, \quad a=2\)
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Fields in Physics

Read Explanation

Force

Read Explanation

Radiation

Read Explanation

Modern Physics

Read Explanation

Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.